Question

If ${\log _5}2$, ${\log _5}\left( {{2^x} - 3} \right)$ and ${\log _5}\left( {\dfrac{{17}}{2} + {2^{x - 1}}} \right)$ are in AP, then the value of $x$ is
A) 0
B) \[-1\]
C) 3
D) None of these

Answer 1

Hint:
Here we need to find the value of the variable used in the question. Here we will use the property and the formula of arithmetic series. Then we will use the property of the logarithm function. After simplifying and using all these properties, we will get the required value of the variable.

Formula used:
Some properties of logarithm function are:-
1) ${\log _a}b - {\log _a}c = {\log _a}\left( {\dfrac{b}{c}} \right)$
2) If ${\log _a}x = {\log _a}y$, then $x = y$.

Complete step by step solution:
Three terms of an AP are given.
 ${\log _5}2$, ${\log _5}\left( {{2^x} - 3} \right)$ and ${\log _5}\left( {\dfrac{{17}}{2} + {2^{x - 1}}} \right)$.
We know that if the numbers are in AP then the difference between any two consecutive terms is constant.
So we can say that the difference between first and the second and the difference between the second and the third term are equal.
Mathematically, we can it as;
$ {\log _5}\left( {{2^x} - 3} \right) - {\log _5}2 = {\log _5}\left( {\dfrac{{17}}{2} + {2^{x - 1}}} \right) - {\log _5}\left( {{2^x} - 3} \right)$
We know from the properties of logarithm function that ${\log _a}b - {\log _a}c = {\log _a}\left( {\dfrac{b}{c}} \right)$
Using this property in the above equation, we get
$ \Rightarrow {\log _5}\left( {\dfrac{{{2^x} - 3}}{2}} \right) = {\log _5}\left( {\dfrac{{\dfrac{{17}}{2} + {2^{x - 1}}}}{{{2^x} - 3}}} \right)$
We also know from the properties of the logarithm function that if ${\log _a}x = {\log _a}y$, then $x = y$.
Using this property in the above equation, we get
$ \Rightarrow \dfrac{{{2^x} - 3}}{2} = \dfrac{{\dfrac{{17}}{2} + {2^{x - 1}}}}{{{2^x} - 3}}$
On further simplifying the terms, we get
$ \Rightarrow \dfrac{{{2^x} - 3}}{2} = \dfrac{{17 + 2 \times {2^{x - 1}}}}{{2 \times \left( {{2^x} - 3} \right)}}$
On further simplification, we get
$ \Rightarrow {2^x} - 3 = \dfrac{{17 + 2 \times {2^{x - 1}}}}{{\left( {{2^x} - 3} \right)}}$
On cross multiplying the terms, we get
$ \Rightarrow {\left( {{2^x} - 3} \right)^2} = 17 + 2 \times {2^{x - 1}}$
We know the algebraic identity that ${{\left( a-b \right)}^{2}}=a{}^\text{2}+b{}^\text{2}-2\cdot a\cdot b$
Using this identity here, we get
$ \Rightarrow {\left( {{2^x}} \right)^2} + {3^2} - 2 \times {2^x} \times 3 = 17 + 2 \times {2^{x - 1}}$
Using the properties of exponents ${{x}^{a}}\times {{x}^{b}}={{x}^{a+b}}$, we get
$\Rightarrow {{\left( {{2}^{x}} \right)}^{2}}+9-{{2}^{x}}\times 6=17+{{2}^{x}}$
On adding and subtracting the like terms of both the sides, we get
$\Rightarrow {{\left( {{2}^{x}} \right)}^{2}}-7\times {{2}^{x}}-8=0$
Let ${2^x} = t$
Now, we will substitute this value here.
$ \Rightarrow {t^2} - 7 \times t - 8 = 0$
Now, we will factorize the quadratic equation that is formed.
$\Rightarrow {{t}^{2}}-8t+t-8=0$
On further simplifying the terms, we get
$ \Rightarrow t\left( {t - 8} \right) + \left( {t - 8} \right) = 0$
Taking common the factors from the two terms, we get
 $ \Rightarrow \left( {t + 1} \right)\left( {t - 8} \right) = 0$
This is possible when $t = - 1$ or $t = 8$
We know that ${2^x} = t$, so we can write it as
${2^x} = 8$ but ${{2}^{x}}\ne -1$
We can write ${{2}^{x}}=8$ as ${{2}^{x}}={{2}^{3}}$
We know from the properties of exponent that if ${{x}^{a}}={{x}^{b}}$, then $a=b$
Using this property in the above equation, we get
$ x = 3$
Hence, the required value of $x$ is equal to 3.

Thus, the correct option is option C.

Note:
Here we have used the properties of logarithm function and exponents. A logarithmic function is defined as the function which is an inverse function of the exponential function. Here, we need to remember that the value of any logarithmic function can’t be negative.
Here, the series is in AP i.e. Arithmetic Progression. An arithmetic progression is a series or sequence in which the consecutive terms differ by a common difference.